# An ionic compound AB has a fluorite-type structure. If the radius of B is 200 pm, what is the ideal radius of A? Options: A: 82.8 pm B: 146.4 pm C: 40 pm D: 45 pm

## Question

An ionic compound $AB$ has a fluorite-type structure. If the radius of $B$ is $200$ pm, what is the ideal radius of $A$?

Options:

- A: $82.8$ pm
- B: $146.4$ pm
- C: $40$ pm
- D: $45$ pm

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## Answer

The correct answer is: **D**

The solution can be understood as follows:

In an **ideal fluorite structure**, the cation ($A$) is present in a **tetrahedral void**. Hence, we can use the relationship for the ratio of ionic radii in this structure:

$$ \frac{r_c}{r_a} = 0.225 $$

Given that the radius of $B$ ($r_c$) is $200$ pm, we can solve for the radius of $A$ ($r_a$):

$$ r_a = \frac{r_c}{0.225} = \frac{200 \text{ pm}}{0.225} \approx 44.4 \text{ pm} $$

Since the closest option to this calculation is **45 pm**, the ideal radius of $A$ is:

**Option D: 45 pm**

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